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Let's say there's a static fluid in an arbitrarily deep vessel with constant non-zero compressibility ($k=\frac{1}{B}=-\frac{dV}{Vdp}=\frac{d\rho}{\rho dp}$). Let $z$ be the vertical distance below the surface of the fluid. Then $dp=g\rho dz$ ($g$ here is positive and assumed to be constant). As $k$ is constant, $\rho$ can be expressed in terms of $p$: $\rho=C_{0}e^{kp}$. If $\rho=\rho_{0}$ and $p=p_{0}$ at $z=0$, then $C_{0}=\rho_{0}e^{-kp_{0}}$, which is a positive quantity. So $e^{-kp}dp=C_{0}gdz$, which can be solved to get $p(z)=-\frac{1}{k}\ln(C_{1}k-C_{0}kgz)$. This doesn't make sense, because the coefficient of $z$ is negative, so there exists a positive value of $z$ for which $p=\infty$. What did I do wrong?

Joshua Meyers
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You cannot write $ \large \frac {dV}{V} = - \large \frac {d\rho}{\rho} $

Your are not allowed to do this because you cannot write $V(z) * \rho(z) = Constant$

Of course, you think that the total mass would be this constant, but that is not correct, in fact you have :

$$ Mass(0 \rightarrow z) = \int_0^z ~ \rho(z) ~ dV(z)$$

[EDIT] A possible solution to your problem is to consider that the relative modification of density would be proportionnal to the pressure, that is:

$$ \frac{\rho(z) - \rho_0}{\rho_0} = k p(z)$$ where $\rho_0$ is the density for $k=0$

Replacing $\rho$ by its value in the equation $dp(z) = g \rho(z) dz$ would give you :

$$dp(z) = g \rho_0 (1 + kp(z)) dz$$

The solution of this equation would be :

$$p(z) = \frac{e^{kg \rho_0 z} - 1}{k} $$

The expression for the density would be :

$$\rho(z) =\rho_0 e^{kg \rho_0 z}$$

You can check, that, when $k \rightarrow 0$, we recover the correct values.

Trimok
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  • How is $\int_{0}^{z}{\rho(z)dV(z)}$ constant? Wouldn't it depend on the value of $z$?

    You're right that I can't write $\rho(z)V(z)=Constant$, but couldn't I write $\rho\Delta V=Constant$, and then apply it the same way, if $k=\frac{d\Delta V}{\Delta Vdp}$?

     – Joshua Meyers Jun 08 '13 at 14:49
  • You are right. Sorry. This mass is only the mass between the coordinates 0 and z (it is better to suppose a surface S constant, for instance). But it did not infirm the first caveat. I made an edit. – Trimok Jun 08 '13 at 18:18
  • From my point of view, Your $\Delta V$ is in fact a $dV$, sot it is a infinitesimal (differential) operator, and it could not be a constant. – Trimok Jun 08 '13 at 18:24
  • In fact, if you have a maximum depth $z_0$, you have a constant which is $Mass(0 \rightarrow z_0)$, so maybe you may try to use this constant to do something. – Trimok Jun 08 '13 at 18:31
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    You're right, an infinitesimal can't be constant. But couldn't you get around that by, considering the set of well-behaved regions with mass $\Delta m$, saying that $\rho \Delta V$ approaches being constant over this set as $\Delta m$ approaches $0$? – Joshua Meyers Jun 08 '13 at 23:34
  • @JoshuaMeyers : I made an edit to propose a solution – Trimok Jun 09 '13 at 09:04

  • I think you're using $p$ as the gauge pressure. Is this correct?

  • I followed your steps and got $p(z)=\frac{e^g\rho_{0}z-1}{k}$. (no $k$ in exponential). Then $\rho=p_{0}e^{g\rho_{0}z}$, which, interestingly, doesn't contain $k$.

  • Do you mean $k$ to be the compressibility, or just an arbitrary constant?

  • I know there are other ways to do this problem (such as the way you proposed), but I still am confused as to why the way I did doesn't work. Looking at it now, there's something weird about $\rho=C_{0}e^{kp}$. Maybe the assumption of constant compressibility is a problem.

    • – Joshua Meyers Jun 09 '13 at 11:21
  • Yes $p$ is the pressure. You certainly made an error in your steps, because $g$ is an acceleration and $g \rho_0 z$ has the dimension of the pressure, so these terms cannot be a non-dimensional factor of exponentials. The $k$ I use has the dimension of compressibility, so you could consider it as a compressibility. In my proposition, it it defined by $\frac{\rho(z) - \rho_0}{\rho_0 p(z)} = k $ – Trimok Jun 10 '13 at 07:24
  • Yeah, I made a mistake. You're right, it's $p(z)=\frac{e^{kg\rho_{0}z}-1}{k}$. But I still can't see why my way didn't work, do you know? – Joshua Meyers Jun 10 '13 at 11:20
  • Maybe the problem is that, in fact, there are 2 variables $z$ and $k$. So the pressure and the density are function of these $2$ variables. So, one have to be cautious about using total derivatives, when there are, in fact, 2 variables. – Trimok Jun 10 '13 at 11:34
  • I was assuming that $k$ was constant. Do you think this is the problem? – Joshua Meyers Jun 11 '13 at 03:05
  • Fundamentally $k$ is constant, and, in my proposal, this is the case. But, Well, I am trying to understand what's wrong with you initial proposal.. My feeling is that, if we want to use your formalism, at some step, it is possible that we have to use 2 distincts degrees of freedom. The first one is, of course, the height $z$. So, maybe, with your formalism, we have to consider $k$ as an other degree of freedom. If my hypothesis is correct, maybe you will have to use partial derivatives instead of total derivatives. Maybe my hypothesis is wrong, but unfortunately I have nothing better. Sorry. – Trimok Jun 11 '13 at 05:47
  • Note that there are 2 kinds of compressibility. Isothermal compressibility (Temperature= Constant, slow change of density) and adiabatic compressibility (Entropy = Constant, fast change of density). – Trimok Jun 11 '13 at 05:56
  • In this reference, you have an interesting graph (Compressibility of water) which shows a dependance of water compressibility with temperature. – Trimok Jun 11 '13 at 06:26